Mathematics Paper 1 Questions and Answers - Form 3 End Term 3 Exams 2021

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INSTRUCTIONS TO THE CANDIDATES

  • This paper contains two sections; Section 1 and Section 11.
  • Answer all the questions in section 1 and only five questions from Section 11
  • All workings and answers must be written.
  • Marks may be given for correct working even if the answer is wrong.
  • Calculations and KNEC Mathematical tables may be used EXCEPT where stated otherwise.
  • Show all the steps in your calculations, giving your answers at each stage.


 QUESTIONS

SECTION I (50 MARKS) 
Answer all questions in this section in the spaces provided.

  1. Without using a calculator ,evaluate
        F3 querry 1
    Giving your answer as mixed fraction (3mks)
  2. Two boys and a girl shared some money. The younger boy got 5/18 of it; the elder boy got 7/12 of the remainder and the girl got the rest. Find the percentage share of the younger boy to the girl’s share. (4mks)
  3. Three numbers, 1400,1960 and n have a G.C.Dand L.C.M of 70 and 22 x 52 x 72 x 11 respectively.Find the least possible value of n (3mks)
  4. A bus starts off from Kitale at 9a.m and travels towards Kakamega at a speed of 60km/hr. At 9.50a.m, a matatu leaves Kakamega and travels towards Kitale at a speed of 60Km/h. How far from Kitale will the two vehicles meet? ( 3mks)
  5. Find the equation of a straight line which is equidistant from the points A(2,3) and B (6,1) (3mks)
  6. Simplify the expression completely (3mks)
    Expression to simplify
  7. Given that sin θ =2/3 and θ is an acute angle, find without using tables tan2 θ + Cos2 θ. Give your answer as a mixed fraction. (3mks)
  8. Solve for y in the equation below. (4mks)
    8(22)y = 6 (2y) – 1
  9. Using a ruler, a pair of compasses only and (proportional) a set square, construct on the upper side division of line BC, a line BD such that ∠DBC = 37.50. Use the line BD to divide BC into 4 equal portions. (3mks)
  10. Sketch the net of the solid below. (2mks)
       Net solid
  11. In a regular polygon, each interior angle is x0 and each exterior angle is regular polygon angles
    1. Find angle X0 (1mk)
    2. Find the number of sides of the polygon (2mks)
  12. The figure below represents a plot of land ABCD such that AB= 85m, BC 75m CD= 60m DA = 50m and angle ACB = 900. (not drawn to scale)
          figure representing plot of land
    Determine the area of the plot, in hectares correct to two decimal places. (4mks)
  13. An open rectangular box measures externally 32cm long, 27cm wide and 15cm deep. The box is made up of metal 1cm thick. If it has a mass of 1.5kg, what is the density of the box to 4 significant figures? (3mks)
  14. Find the integral values of x which satisfy the following inequalities;
    2x + 3 > 5x – 3 > -8 (3mks)
  15. A Kenyan bank buys and sells foreign currency as shown below.
        Buying Ksh          Selling Ksh    
    1 US dollar ($)   63.00  63.20
     1 UK pound (£)  125.00  125.95
    A tourist arrived in Kenya with £ 9600 which he converted into Kshs at a commission of 5%. He later used ¾ of the money before changing the balance of dollars at no commission calculate ; to the nearest dollar, the amount he received. (3mks)
    The histogram shown below represents the distribution of marks obtained in attest. The bar marked A has a height of 3.2 units while B has aheight 1.2 units. If the frequency of the class represented by B is 6, find the frequency of the bar represented by A. (3mks )

          histogram pic
SECTION II (50 MARKS)
Answer any five questions in this sections in the spaces provided.

  1. The figure below (not drawn to scale) shows a quadrilateral ABCD inscribed in a circle. AB = 5cm, BC = 8cm, CD = 7cm and AD = 8cm. AC is one of the diagonals of length 10cm.
             Quadrilateral inscribed in a circle
    1. Find the size of angle ABC. (3mks)
    2. Find the radius of the circle. (2mks)
    3. Hence, calculate the area of the shaded region. (5mks)
  2. In the figure below     →           →      →          →
                                    OB = b , OC = 3OB and OA = a
                             →            →        →          →   →         →         
    1. Given that OD = 1/3 OA and AN = ½ AC, CD and AB meet at M. Determine in terms of a and b
              triangle fig 18
                →                                       
      1. AB (1mk)
         →
      2. CD (1mk)
    2.                  →        →         →         →
      Given that CM = k CD and AM = h AB determine the values of the scalars k and h (5mks)
    3. Show that O,M and N are collinear. (3mks)
  3. Three variables p, q and r are such that p varies directly as q and inversely as the square of r.
    1. When p = 18, q = 24 and r = 4.
      Find p when q = 30 and r = 10. (4mks)
    2. Express q in terms of p and r. (1mk)
    3. If p is increased by 20% and r is decreased by 10% find:
      1. A simplified expression for the change in q in terms of p and r. (3mks)
      2. The percentage change in q. (3mks)
  4. A circular lawn is surrounded by a path of uniform width of 7m. The area of the path is 21% that of the lawn.
    1. Calculate the radius of the lawn. (4 marks)
    2. Given further that the path surrounding the lawn is fenced on both sides by barbed wire on posts at intervals of 10 metres and 11 metres on the inner and outer sides respectively. Calculate the total number of posts required for the fence. (4 marks)
    3. Calculate the total cost of the posts if one post costs sh 105. (2 marks)
  5. The diagram below represents two vertical watch – towers AB and CD on a level ground. P and Qare two points on a straight road BD. The height of the tower AB is 20m and road BD is 200m
            diagram of two vertical watch
    1. A car moves from B towards D. At point P, the angle of depression of the car from pointsA is 11.30Calculate the distance BP to 4 significant figures. (2mks)
    2. If the car takes 5 second to move from P to Q at an average speed of 36km/hr, calculate the angle of depression of Q from A to 2 decimal places. (3mks)
    3. Given that QC = 50.9 cm, calculate
      1. The height of CD in meters to 2 decimal places; (2mks )
      2. The angle of elevation of A from C to the nearest degree. ( 3mks)
  6. The parents of acertain mixed secondary school decided to buy a school van worth Ksh. 900,000. Each student was to contribute the same amount of money. 50 students were transferred from the school; as a result each f the remaining students had to pay Ksh. 600 more.
    1. Find the original number of the students in the school. (5mks)
    2. Find the percentage change in contributions per student. (3mks)
    3. If the ratio of boys to girl in the school was 11:7 find the amount money contributed by boys alone.(2mks)
  7. Five members of ‘SILK’, a self-supporting enterprise Jane, Jepchoge, Esther, Mama Charo and Chepkoech were given a certain amount of money to share amongst themselves. Jane got 3/8 of the total amount while Jepchoge got 2/5 of the remainder. The remaining amount was shared equally among Esther, Mama Charo and Chepkoech each of which received Kshs. 6,000;
    1. How much was shared among the five business women? (3mks)
    2. How much did Jepchoge get? (2mks)
    3. Jane, Jepchoge and Chepkoech invested their money and earned a profit of Kshs. 12,000. A third of the profit was left to maintain the business and the rest was shared according to their investments. Find how much each got. (5mks)
  8. Onyango and Juma live 200km apart. One day, Onyango left his house at 7.00am and travelled towards Juma’s house at an average speed of 30km/hr. Juma left his house at 7.30am on the same day and travelled towards Onyango’s at an average speed of 40km/hr.
    1. Determine:-
      1. The time they met. (2mks)
      2. The distance from Onyango’s house where the two met. (2mks)
      3. How far was Onyango from Juma’s house when they met? (2mks)
    2. The two took 15 minutes at the meeting point and then travelled to Juma’s house at an average speed of 20km/hr. Find the time he arrived at Juma’s house.(2mks)


MARKING SCHEME

SECTION 1

  1. 3/4 + 15/7 ÷ 4/7 × 7/3
    = 3 + 9 = 21 + 36 = 57
       4    7          28       28
    = (80 − 35)× 2
             56        3
    =45/56 × 2/3
    =15/28
    =57/28 × 28/1519/5
    = 34/5
  2. 7/12 × 13/18 =91/216
    91  ×  = 91 + 60 = 151
        216     18       216        216
     (5/18 ÷ 216/65)100
    =× 216 × 100 = 1200
       18    65                 13
    = 924/13%
  3.   1400 
     2 
     2
     2
     5
     5
     7 
     700
     350
     175
      35
       7
       1

      1960 
     2 
     2
     2
     5
     7
     7  
     980
     490
     245
     47
      7
       1
    L.C.M =23 × 5× 11
          N = 23 × 5 × 72
        70 
     2 
     5 
     35
      7
    L.C.M
    23 × 5× 72 × 11
    n = 2× 5 × 11 × 7
       = 770
  4. At 9.50am , the bus has travelled
    (20/60 × 60) = 20km
    The distance between the two vehicles at 9.50am
    (65 − 45) =20km
    Rel . speed = 140 km/ h.
    It will take them 45/140hrs since leaving Kitale.
    Distance covered at the time Matatu met the bus
    =45/140 × 80 = 25.71km

  5.   
    q5 form 3
    Gradient = 1−3 = −1
                     6−2     2
     g × −1/2 =−1
                 g=2
            y−2 =2
            x−4
            y−2=2x−8
                y=2x −6
  6.      4x(3x−4)       =       4x(3x−4)         
    20−15x+4x−3x2      5(4−3x) + x(4−3x)
                              =     4x(3x−4)   
                                  (5+x)(4−3x)
                              = −4x(4−3x)  
                                 (5+x)(4−3x)
                              = −4x
                                 5+x
  7.  
       traingle image       
       y=√9−4
        =√5
       tan2 θ + Cos2 θ
       (  /√5)2 + ( √5/3)2
       = 4 + 5 = 61
          5     9    45
       =116/45
  8. 8x(22)y = 6(2y)−1
    =8x(2y)2 = 6x(2y)−1
     Let 2y = x
    ∴ 8x2 = 6x−1
    →8x2−6x+1=0
    →8x2−4x−2x+1=0
       4x(2x−1)−1(2x−1)=0
    →(4x−1)(2x−1)=0
    →∴x = ½ or ¼
        2y=½ = 2−1  or   2y =¼ = 2−2
        ∴y=−1
        or
          y=−2
  9.  
       q10 ruler compass set square
  10.  
       net solid sketch
  11. x + x−36 =1800
            3
    →3x + x−36 =540
                   4x =576
                   →x= 1440
    Exterior angle= 144 − 36 =108 =360
                                  3          3
                    ∴No of sides of the polygon
                          n= 360 =10sides
                                36
  12. AC2= 852 − 752
     AC =SQUARE ROOT 11=40m
    Area of quadrilateral ABCD
    ½ × 40 × 75 + sqaure root of
          =1500 + √984375
          =2492m2
    In hectares = 2492   = 0.2492hectares
                        10000
                     ≈0.25hectares
  13.  
         open rectangular box
    External volume =(32 × 27 × 15) =12960cm3
    Internal volume =(30 × 25 × 14) =10500cm3
    Volume of material = (12960 − 10500)
                    Metre used = 2460cm3
                 Density =  mass   = 1500
                                volume     2460
                             =0.6098g/cm3
  14. 2x>5x−3>−8
    2x+3>5x−3
      −3x>6
          x<2
    5x−3>−8
         5x>−5
           x>−1
        −1<x<2
    The integral values are 0 and 1
  15. Remaining amount = 9600 × 95 × 125
                                      1E    100
                                =Ksh 1,140,000
          Amount speed = ksh 1,140,000 × 3  
                                                    4
              The Balance = ksh 1,140,000- 855,000
                                 =ksh285,000
       Amount in dollars = 285,000 
                                      63.20
                                 =4509.49 USdollars

  16. 1.2 × 10 × x = 6
                     x = 6/12
                        =0.5
    3.2 × 15 × 0.5 = 24

 SECTION II

  1.  
                 Shaded circle        

    1. 102 =82 + 5− 2 × 8 × 5 Cos B
      Cos B=89 − 100 = −11
                     80           80
            B = Cos−1(−11/80)=97.900
    2. 2R =      10      
              Sin 97.90
        R =       5        
               Sin 97.90
           =5.0479cm
    3. Sin A = 7/10 Sin 82.1 = 0.6984
            A = Sin−1(0.6934) = 43.900
      ∠COD = 2 × 43.900= 87.800
        Area =87.80  × 22  × 5.0479 − ½ ×5.04329 Sin 87.80
                    360       7   
               = 19.5316 − 12.7313
               = 6.800cm2
  2.  
    1.  

      1. AB = ba

      2. CD = −3b1/3a
    2. CM = 1/3K a − 3b
      →AM = −h a + h b
         CM = CA+AM
            =1/3K a − 3 b
             = a(1− h) + b(h−3)
      K=3−3h and 1/3K =1−h
      h=¾
      k=¾
    3.  →
      OM=¼a + ¾b
       →
      ON =½a + ¾b
       →         →
      OM = ½ON
      Hence parallel, and point O is common
      → O, M and N are collinear
  3.  
    1. p ∞ q/r
      p =kq/r2
      18 =24 × k 
                 42
      K= 18 × 16 
                2
      K=12
      Equation: p=12q/r2 
      when q=30, r=10
      P=(12 × 30)
               100
        = 18/5 or 3.6
    2. P= 12q/r2 
      12q= pr2
         q = pr2/12 
    3.  
      1. New value of P =1.2P
        New r = 0.9r
        q =pr2/12 
          =1.2p × (0.9r)2
                   12

        Δin q = variables q19
        = pr2(1.2 × 0.81 – 1)
           12
        =–0.028 pr2
               12
      2. –0.028 pr2 × 12 × 100
            12             pr
        =2.8%
  4.  
    1. R = r + 7
      πR2 − πr2 21  πr2   
                        100
      πR2=0.21 πr2 + πr2
      πR= 1.21r2
        R2 = 1.21r2
         R =1.2r
      R + 7 = 1.1r
           7= 1.1r−r
           7= 0.1r
           R=70m
    2. Inner radius =70m
      2πr = 2 × 22× 70 =400m
                      7
      440m = 44 Posts
       10
      Outer radius r=77
      2πr = 2 × 22× 77 =484m
                      7
      484  = 44posts
       11
      Total number of posts =88
    3. Total cost
      88 × 105 = Ksh 9240
  5.  
        vertical watch towers
    1. Tan 11.30 =20
                       BP
      ∴BP =     20       = 100.09m
               Tan 11.30
    2. Time taken in hrs =  5   
                                  3600
      Distance netween P and Q
      =  equation 21
      =0.05km
      =50km
      → tan α =   20    
                     150.09
      α = tan−1 (  20    ) = 7.590
                     (150.09)
    3.  
      1. CD2 = 50.92 − 49.912 = 99.8019
         CD = √99.8019 =9.9900
               = 9.99m
      2. tan α = 10.01 = 0.05005
                     200
              α = tan−1(0.05005)
                 = 2.865
                 =3.00   

    1. 900,000 − 900,000 = 600
      x − 50           x
       1500  1500  =1
      x −50        x
      1500x − 1500x + 75000
               x(x − 50)
      75000 =x2 − 50x
      x2 − 50x −75000 = 0
      x2 −300x + 250x −75000 =0
      x(x − 300) + 250(x − 300) =0
      (x − 300)(x + 250) =0
      →x=300
    2. 900,000  900,000 × 100%
          250           300
       = (3600 − 3000) × 100%
          (      3000      )
       = 20%
    3. B : G
      11:7
      11 × 900,000
      18
      =Kshs 550,000
  6.  
    1. Let the constant amount be x
      Jane-(3/8x)/=
      Jepchoge's- 2/5(5/8x)/=
                    =¼x/=
      Remaining 3/5x − 18,000
                  x=48,000/=
      Therefore the original amount is 48,000
    2. Jepchoge received
      (¼ × 48,000)
      = 12,000/=
    3. Business maintenance
      =(1/3 × 12,000)
      = 4,000/=
      Balance = 8,000/=
      Ratios: Jane=(3/8 × 48,000)
                        = Kshs. 18,000
           Jepchoge=(¼ × 48,000)
                         =Kshs. 12,000
       Chepkoech=1/3 × 18,000)
                          =Kshs. 6,000
      Ratio: 18,000 : 12,000 : 6,000
                   3      :     2     :     1
      Jane got 3/6 × 8,000  
                      =Kshs. 4,000
      Jepchoge got 2/6 × 8,000
                      =Kshs. 2,667
      Chepkoech got 1/6 × 8,000
                       =Kshs.1,333
  7.  
    1.  

      1. relative speed

        R.S = 30km/hr + 40km/hr
              = 70km/hr √m1
        D= S×T
          = 30km/hr ×½hr
          =15km
        T =
             R.S
          185km  
             70km/hr
        = 2 ½ hours  √m1
        Time they met = 7.30am √m1
                               +2.30
                              = 10.00am √m1

      2. Distance from Onyango’s house where the two met:
        Distance = 30km/hr x 5/2 hours
        = 75km
        = 75 + 15
        = 90km √m1

      3. Distance from Onyango to Juma’s house when they met:
        200km – 90km √m1
        = 110km √A1

    2.  Meeting time = 10.00a.m
                               + 15
                              10.15am
      T =   110km   = 5 ½ hours√m1
             20km/hr
      Time of arrival to Jumas house
      = 10.15am
         +5.30√m1
      = 15.45pm or 3.45pm

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